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← JEE Main 2014 · 11 Apr · Shift 1
MathematicsSingle correct+4 / -1Question 65 of 90

JEE Main 2014 · 11 Apr · Shift 1, Mathematics question 65

Let for i=1,2,3i=1,2,3, pi(x)p_i(x) be a polynomial of degree 2 in xx, pi′(x)p_i'(x) and pi′′(x)p_i''(x) be the first and second order derivatives of pi(x)p_i(x) respectively. Let,
A(x)=[p1(x)p1′(x)p1′′(x)p2(x)p2′(x)p2′′(x)p3(x)p3′(x)p3′′(x)]A(x)=\begin{bmatrix}p_1(x)&p_1'(x)&p_1''(x)\\p_2(x)&p_2'(x)&p_2''(x)\\p_3(x)&p_3'(x)&p_3''(x)\end{bmatrix}
and B(x)=[A(x)]TA(x)B(x)=[A(x)]^TA(x). Then determinant of B(x)B(x) :

Choose one option
Answer and worked solution

Official answer: D

With pi=aix2+bix+cip_i=a_ix^2+b_ix+c_i, apply the column operation C1→C1−xC2+x22C3C_1\to C_1-xC_2+\frac{x^2}2C_3. The first column becomes (ci)(c_i) and the second (2aix+bi)(2a_ix+b_i); then C2→C2−xC3C_2\to C_2-xC_3 leaves (bi)(b_i), with C3=(2ai)C_3=(2a_i).

So det⁡A\det A is constant, and det⁡B=(det⁡A)2\det B=(\det A)^2 does not depend on xx.